Question

Amongst the following, the total number of compounds soluble in concentrated $N{H}_3$ solution is:
$A{g}_{2}Cr{O}_4,Cu(OH{)}_2,PbS{O}_3,Al(OH{)}_3,Ni(OH{)}_2,Z{n}_3(P{O}_4{)}_2,BaS{O}_4$

• A. $4$
• B. $7$
• C. $5$
• D. $6$

Answer 1

Answer: (A)

$4$

(A) $A{g}_{2}Cr{O}_4\left(yellow\right)+N{H}_3\to {\left[Ag{\left(N{H}_3\right)}_2\right]}^{+}+{Cr{O}_4}^{2-}$
(B) $Cu\left(OH{)}_2\right(Blue)+8N{H}_3\to 2{\left[Cu{\left(N{H}_3\right)}_4\right]}^{2+}+{S{O}_4}^{2-}+2OH-$
(C) $PbS{O}_3\left(White\right)+4O{H}^{-}\to \left[Pb{\left(O{H}_4\right)}^{2-}\right]+{S{O}_3}^{2-}$

Soluble in sodium hydroxide not in NH${}_3$.
(D) $Al(OH{)}_3$ is soluble in sodium hydroxide not in NH${}_3$.
(E) $Ni{\left(OH\right)}_2\left(green\right)+6N{H}_3\to {\left[Ni{\left(N{H}_3\right)}_6\right]}^{2+}$(deep blue colouration) $+2O{H}^{-}$
(F) $Z{n}_3{\left(PO\right)}_4\left(white\right)+12N{H}_3\to 3{\left[Zn{\left(N{H}_3\right)}_4\right]}^{2+}+2{P{O}_4}^{3-}$
(G) $BaS{O}_4$ is insoluble in $N{H}_3$, soluble appreciably in boiling concentrated $H_{2}S{O}_4$.